Solve {l}{2003x-2004y=4}{2005x-2006y=6}

$$\left\{ \begin{array} { l } { 2003 x - 2004 y = 4 } \\ { 2005 x - 2006 y = 6 } \end{array} \right.$$

Solve for x, y

$$x=2000$$

$$y=1999$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$2003x-2004y=4,2005x-2006y=6$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2003x-2004y=4$$

Add 2004y to both sides of the equation.

$$2003x=2004y+4$$

Divide both sides by 2003.

$$x=\frac{1}{2003}\left(2004y+4\right)$$

Multiply \frac{1}{2003} times 2004y+4.

$$x=\frac{2004}{2003}y+\frac{4}{2003}$$

Substitute \frac{2004y+4}{2003} for x in the other equation, 2005x-2006y=6.

$$2005\left(\frac{2004}{2003}y+\frac{4}{2003}\right)-2006y=6$$

Multiply 2005 times \frac{2004y+4}{2003}.

$$\frac{4018020}{2003}y+\frac{8020}{2003}-2006y=6$$

Add \frac{4018020y}{2003} to -2006y.

$$\frac{2}{2003}y+\frac{8020}{2003}=6$$

Subtract \frac{8020}{2003} from both sides of the equation.

$$\frac{2}{2003}y=\frac{3998}{2003}$$

Divide both sides of the equation by \frac{2}{2003}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1999$$

Substitute 1999 for y in x=\frac{2004}{2003}y+\frac{4}{2003}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2004}{2003}\times 1999+\frac{4}{2003}$$

Multiply \frac{2004}{2003} times 1999.

$$x=\frac{4005996+4}{2003}$$

Add \frac{4}{2003} to \frac{4005996}{2003} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=2000$$

The system is now solved.

$$x=2000,y=1999$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$2003x-2004y=4,2005x-2006y=6$$

Write the equations in matrix form.

$$\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\6\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right).

$$inverse(\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right))\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right))\left(\begin{matrix}4\\6\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right))\left(\begin{matrix}4\\6\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2003&-2004\\2005&-2006\end{matrix}\right))\left(\begin{matrix}4\\6\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2006}{2003\left(-2006\right)-\left(-2004\times 2005\right)}&-\frac{-2004}{2003\left(-2006\right)-\left(-2004\times 2005\right)}\\-\frac{2005}{2003\left(-2006\right)-\left(-2004\times 2005\right)}&\frac{2003}{2003\left(-2006\right)-\left(-2004\times 2005\right)}\end{matrix}\right)\left(\begin{matrix}4\\6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1003&1002\\-\frac{2005}{2}&\frac{2003}{2}\end{matrix}\right)\left(\begin{matrix}4\\6\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1003\times 4+1002\times 6\\-\frac{2005}{2}\times 4+\frac{2003}{2}\times 6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\1999\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2000,y=1999$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2003x-2004y=4,2005x-2006y=6$$

To make 2003x and 2005x equal, multiply all terms on each side of the first equation by 2005 and all terms on each side of the second by 2003.

$$2005\times 2003x+2005\left(-2004\right)y=2005\times 4,2003\times 2005x+2003\left(-2006\right)y=2003\times 6$$

Simplify.

$$4016015x-4018020y=8020,4016015x-4018018y=12018$$

Subtract 4016015x-4018018y=12018 from 4016015x-4018020y=8020 by subtracting like terms on each side of the equal sign.

$$4016015x-4016015x-4018020y+4018018y=8020-12018$$

Add 4016015x to -4016015x. Terms 4016015x and -4016015x cancel out, leaving an equation with only one variable that can be solved.

$$-4018020y+4018018y=8020-12018$$

Add -4018020y to 4018018y.

$$-2y=8020-12018$$

Add 8020 to -12018.

$$-2y=-3998$$

Divide both sides by -2.

$$y=1999$$

Substitute 1999 for y in 2005x-2006y=6. Because the resulting equation contains only one variable, you can solve for x directly.

$$2005x-2006\times 1999=6$$

Multiply -2006 times 1999.

$$2005x-4009994=6$$

Add 4009994 to both sides of the equation.

$$2005x=4010000$$

Divide both sides by 2005.

$$x=2000$$

The system is now solved.

$$x=2000,y=1999$$